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NVAEasyJEE 2023Bernoulli's Theorem

JEE Physics 2023 Question with Solution

Figure below shows a liquid being pushed out of the tube by a piston having area of cross section 2.0 cm22.0 \, \text{cm}^2. The area of cross section at the outlet is 10 mm210 \, \text{mm}^2. If the piston is pushed at a speed of 4 cm/s4 \, \text{cm/s}, the speed of the outgoing fluid is _____ cm/s\text{cm/s}.

A piston pushes liquid through a wide tube that narrows to a small outlet nozzle, with fluid emerging from the right end.

Answer

Correct answer:80

Step-by-step solution

Standard Method

Given: Area at piston A1=2.0 cm2A_1 = 2.0 \, \text{cm}^2, piston speed V1=4 cm/sV_1 = 4 \, \text{cm/s}, outlet area A2=10 mm2A_2 = 10 \, \text{mm}^2.

Find: Speed of the outgoing fluid V2V_2.

Use the equation of continuity:

A1V1=A2V2A_1 V_1 = A_2 V_2

Convert the outlet area into cm2\text{cm}^2:

A2=10 mm2=10×10−2 cm2A_2 = 10 \, \text{mm}^2 = 10 \times 10^{-2} \, \text{cm}^2

Substitute the values:

2×4=(10×10−2)×V22 \times 4 = (10 \times 10^{-2}) \times V_2

So,

V2=2×410×10−2=80 cm/sV_2 = \frac{2 \times 4}{10 \times 10^{-2}} = 80 \, \text{cm/s}

Therefore, the speed of the outgoing fluid is 80 cm/s80 \, \text{cm/s}.

Common mistakes

  • Using the continuity equation incorrectly by assuming velocity stays the same everywhere. This is wrong because the cross-sectional area changes. Use A1V1=A2V2A_1V_1 = A_2V_2 so that smaller area gives larger speed.

  • Not converting 10 mm210 \, \text{mm}^2 into cm2\text{cm}^2 before substitution. This mixes units and gives a wrong answer. First convert area into consistent units, then apply the formula.

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